Susan Brown

[Susan Brown]

Right.

So what your meter will measure is what we call the “equivalent resistance” of those two parallel loads. The equivalent resistance is taking the resistance of loads in parallel and theoretically combining them into a single load.

(FYI – In a series circuit we would calculate the “total resistance”, which is just the sum of the resistances.)

For parallel loads, we described how the “equivalent resistance” can be calculated in Unit 5. We also gave a rule of thumb in one of the videos.

Do you know the rule of thumb? If so, you’ll be able to get the answer to this question correct. Let me know!

[Susan Brown]

We are taking an “ohms” (resistance) measurement.

There are two loads in that circuit: the defrost heater and the drain heater (the bimetal is just a switch).

Are they in series or parallel with each other?

[Susan Brown]

Hi Michael,

There are a few things you need to be able to know to get this answer. I’ll step you through it.

First, do you understand what measurement we are talking about from Pin 13 to Pin 7?

[Susan Brown]

Yes – that is exactly what is happening there!

[Susan Brown]

Hi Thomas,

Sure – I just reset you.

BTW – the best way to ask for a reset is to use our Contact form at the site (“contact” in the main menu).

Although hopefully you’ll have more regular meals so you won’t need a reset again!
🙂

[Susan Brown]

Do you see where the Noise Filter is on the schematic? It shows the line and neutral coming into it, along with the color of the wires.

[Susan Brown]

Yes – just did it. Thanks for the reminder 🙂

[Susan Brown]

HI Matthew,

Great job – those answers are correct!

I hid them so we don’t just give them away to other students.

[Susan Brown]

Yes, this is the right place to ask about the Midterm.

Questions 7 and 8 are about series and parallel circuits:
– being able to recognize which configuration you’ve got in a diagram (are the loads in series with each other or parallel?)
– recognizing special situations like shorts and shunts
– understanding the behavior of current, voltage, and voltage drop in both types of circuits.

That material is covered in units 4, 5, and 8 of Basic Electricity

[Susan Brown]

I’m not sure where you are trying to go. Did you click the link that I gave you above? That goes right to Forum 1.4, which is for Fundamentals, Module 4, Basic Electronics.

[Susan Brown]

Hi Gary,

I’m glad to see you posting a question and am happy to help you, but first we need to get your question clarified and posted in the right place.

The midterm exam is in Module 4, Basic Electronics, so you should post your question in the Forum for Basic Electronics. In fact, if you look at that Forum you’ll see there are already questions about the midterm posted there.

https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/forum/basic-electronics/

Also, the title of your post should clearly reference what you are talking about. This one looks like part of a link to an image, so I don’t know which question you want help with. You can just say something like “Midterm question 1”.

Please do that, and I’ll keep an eye out for that post so I can help you.

[Susan Brown]

Let’s start with this one: I = E/R

The calculation you are referring to is shown in the “Equivalent Resistance…” video in Unit 5.

It is for a series circuit, with 3 loads. In a series circuit, the current is the same throughout the circuit, and the way you calculate it is to add the 3 loads to find the total resistance in the circuit, then use that in the equation above.

Again – these are loads in series. Loads in parallel are treated differently.

So the equation becomes I (current) = E (voltage) divided by R (total circuit resistance)

You need to add the resistances together first, then put that number in the equation. (In the video, he puts the 3 resistance numbers in parentheses. When doing math calculations, you always do the operation shown in parentheses first, then move on to the others.)

What do you come up with?

[Susan Brown]

Hi Matt,

We helped someone else awhile back with calculating equivalent resistance in parallel circuits. Check out this topic and see if that helps you with your first question.
https://my.mastersamuraitech.com/appliance-repair-course-support/student-forums/topic/calculating-equivalent-resistance/

For your second question, I’m not sure if I see what you are showing there. Are you trying to calculate the current (I) when E is 120vac and the total resistances are 58.75 ohms? (those 3 added together). If so, then look at the Ohm’s Law pie chart in Unit 3. If you know what E and R are, and you are trying to find I, what is the formula? (You can just write it in words rather than try to format it)

If I’m misunderstanding your question, please clarify. Thanks!

[Susan Brown]

Hi Matthew,

We like to see students asking questions here, so you didn’t come across in a negative way at all. Feel free to ask questions here about anything you’d like to get help or feedback on.

Yes, your description of current here is exactly what we teach in numerous instructional videos throughout the course, which you’ll see as you progress.

Voltage is what makes the negatively-charged electrons want to move to move towards a relatively more positive charge, and current is the movement of those electrons in the wire.

[Susan Brown]

Hi Matthew,

Sharp eye! Good job looking at the schematic.

That lid switch drawing is very vague and difficult to interpret, and Maytag gives no assistance on the diagram to help us. The symbol that looks like a capital I with bubbles is particularly strange. We have to use our common sense to try to understand what the drawing is showing, but bottom line is we know that the machine is not designed to short out the way it appears to be drawn. In fact, this lid switch assembly has a fuse in it in case a stuck contact creates a short.

Here’s another drawing I found of this lid switch. You can see a few differences, including showing more of a dashed vertical line in that thing I mentioned that looks like an “I”. So that is not a path for current. Hopefully this helps!

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